Optimize Maintenance/Asset Management Policies
EP Editorial Staff | June 1, 2004
Translate contributions into the language of finance.
In the majority of companies today, there is a language barrier. There are at least three different languages spoken internally—financial, operational, and technical.
Just as language can be a barrier in communication with others on a personal level, the different languages spoken in a company also create barriers. For example, operational personnel have difficulty understanding issues faced by the maintenance and engineering departments since they are always expressed in terms such as life cycle costing, mean time between failure, or mean time to repair. Maintenance and engineering personnel may have difficulty understanding cost per unit, return on cost of capital, or total cost to produce.
Language of finance
Ultimately, the one language that must be understood throughout the company is financial. Regardless of the decision being made, the ultimate factor will be expressed in the language of finance. This requires that all other departmental languages translate their contributions into financial terms. However, can this actually be done? Can the maintenance and engineering functions clearly articulate their impact on the financial standing of the company?
Nearly all maintenance and engineering decisions are cost benefit or return on investment decisions. This means that all decisions that involve maintenance should be considered in the light of the financial impact on the company. The maintenance or engineering perspective is typically expressed in extended useful life of assets, increased throughput or availability, or improved product or service quality. Clearly the problem for maintenance and engineering is how to convert these benefits into financial terms so the entire organization can quantify them.
Maintenance impact costs
The majority of maintenance and engineering decisions impact at least five major cost categories:
• Labor
• Material
• Overhead
• Function loss
• Function reduction
Labor costs are the expenses for having maintenance technicians on staff. This will include all related expenses including overtime, benefits, etc. Each job will have the cost of the number of employees times their hourly rate plus any incentives. This cost makes up approximately 50 percent of all direct maintenance expenditures.
The material costs include all parts and equipment replacement expenses controlled by maintenance. The costs of all parts or equipment components used on a job should be tracked through the work order system. Material costs make up the other 50 percent of all direct maintenance expenditures. It should be noted that if maintenance contractors are used in a company, their costs are divided similarly.
Overhead costs include clerical and staff support for maintenance. Typical overhead costs are about 15-25 percent of the usual maintenance labor costs.
The function loss (also known as equipment breakdowns) costs are the hidden or difficult costs to ascertain. Studies have shown that these costs will range from 2:1 to as high as 15:1 per maintenance activity. So when a maintenance task costs $10,000 in labor and materials, function loss costs could run the true total costs of the incident to as high as $30,000-$160,000. This makes a tremendous difference in looking at the cost of a maintenance action.
In addition, there are also function reduction (also known as efficiency losses) costs. In some organizations, these are also known as idling or minor stoppage losses. These occur when the equipment operates, but has intermittent problems that are typically not significant enough to count as a function loss (breakdown) but have a significant impact on the efficiency of the equipment. Studies by Japanese companies have shown that these types of losses always exceed the losses incurred in function loss or equipment breakdowns, sometimes by as much as four times.
The restoration of a function loss breakdown will usually involve some cleaning or replacement of a component on a trend of decreasing efficiency. This could be tasks like cleaning pumps with decreasing efficiency or restoring the full operational capacity to production equipment. This is an overlooked area of maintenance improvement.
Compliance with governmental regulations also can be a function of maintenance. For example, one municipality was fined over $1 million for an environmental violation. The cause of the violation was ultimately traced to a pump that was improperly maintained. Fines are a small part of the total picture when damage to the environment or employees’ health or damage due to improper handling of hazardous materials is included.
Collect the information
It is necessary to put the costs vs benefit discussion in a form that all departments in the company can understand. Figure 1 shows this material in a graphic form, revealing that the decision for maintenance would be made based not on what is best for the operations group, nor on what is best for the maintenance group, but on what is the lowest combined cost. This is the effective “bottom line” for the company and its shareholders. This is the type of decision that companies must make if they are to optimize their resources.
How does a company go about collecting the information required to perform this analysis? It starts with assigning a cost to downtime or unavailability. It may be useful to use the financial or accounting departments to find out what an hour or a shift of lost production is worth for a piece of equipment. This might include lost sales, employee salaries and overhead, the cost to make up lost production (if it can be made up), and any measurable depreciation to the assets. The figures coming from the financial department will usually be conservative, but will not be disputed by other parts of the organization.
With these figures agreed to, it is necessary to understand the maintenance costs involved. This may include the labor, material, or supply and miscellaneous costs that will be incurred due to the repair or the failure. These costs may be needed to compare an overhaul to a run to failure approach to maintenance. Additional costs that may be incurred also should be calculated including hazardous materials, Environmental Protection Agency, Occupational Safety and Health Administration, or safety considerations.
Once the total cost is understood, companies can identify some interesting problems and use this technique to justify and solve them. We will address scenarios in three main areas:
• Preventive maintenance (PM) frequencies
• Critical spares analysis
• Normal maintenance parts and supplies
PM frequencies
To examine PM frequencies, let’s look at a piece of equipment common to most plants or facilities—a centrifugal pump. It may be pumping a product or moving cooling water. The point is it will have a value to its service. Setting a price on the value gives a reference from which to start. If the value is $100 per hour, then this allows managers to solve the following problem:
History shows it cost $1500 for parts and labor to overhaul the pump. There is no downtime cost since there is a standby pump available. The pump performance is measured, and it is found that after 4000 hours of operation, it loses 5 percent of its capacity. In order to simplify the problem, it will be assumed that the drop is linear and continues to be so throughout the life of the pump. When is it cost effective to remove the pump from service and clean the rotor?
The problem is solved by calculating the amount of maintenance cost vs lost performance cost per hour. The two are combined to give the lowest total cost. The techniques to perform this analysis are illustrated in Fig. 2, 3, 4, and 5. In Fig. 2 the maintenance cost is calculated. The mistake of considering only maintenance costs is highlighted in this particular example. If only the maintenance costs were considered, it would be advisable to delay servicing the pump for as long as possible.
The point in Fig. 3 is that if you delay the service, the amount of lost production cost is increasing linearly. However, the difference between Fig. 3 and Fig. 4 is that the latter takes into consideration that the performance fall off is triangular and not the total area volume of the rectangle.
The summary of the problem is in Fig. 5. It plots the falling maintenance cost against the increasing cost of lost performance. These two factors added together will give the total of the true costs. The decision can then be made on lowest true cost, which in this problem would indicate the maintenance action should be performed every 1500 hours of actual run time.
The problem can become more complex, since this example has no penalty cost for the downtime required for the maintenance action. If it would cost additional money for the downtime, then this column would have to be added. However, in the example just given, the only downtime that would likely be incurred is when the repair was made. Some problems will include breakdowns when maintenance intervals exceed a certain level. This means that downtime may have to be factored in during the cycle. This will radically alter the results of the calculation.
For example, continuing to expand the problem in Fig. 5, the downtime costs could be added with the result being Fig. 6. Now the downtime drives the true cost even higher than it was previously. However, in this example, the initial cost of the downtime is factored in immediately. The breakdown will occur only if maintenance is not performed before 3000 hours of operation. If the maintenance frequency extends beyond that time, an additional cost of $2400 (24 hr x $100/hr value of the process) will be incurred. This removes any doubt that the lowest total cost would be around 2000 hours of operation, but definitely before 3000 hours of operation.
Useful statistical techniques
The ability to apply statistical techniques can go far beyond the simple example used in this article. Consider the ability to perform this type of costing for each subcomponent of a large mechanical drive; the ability to determine the lowest life cycle costing for complex equipment; the ability to determine the amount of resources to be spent on redesign and retrofit engineering projects, based on anticipated return on investment.
However, while this is a straightforward mathematical model, why are there so few companies doing this type of calculations? It is because few of them have any reliable data with which to work. This clearly highlights the necessity to have a good computerized maintenance management/enterprise asset management (CMMS/EAM) system in place with accurate historical records. The data to perform these calculations must come from the work order system. Without this reliable data, companies will be back to guessing when maintenance activities should take place and not financially justifying the maintenance activity.
Critical spares analysis
A second example of statistical financial controls examines critical maintenance spares. Spare parts are an albatross for maintenance managers. Operations and upper management see critical spares (which are typically slow moving) as unnecessary since they never seem to turn over. The same techniques can be applied to the spares as was applied to the equipment. For example, consider what costs must go into a stocking decision:
• How many are used per year or mean time between failures?
• What are the operating requirements for the equipment per year?
• What is the cost of downtime if the part is unavailable?
• What is the annual holding cost for the item?
• What does the item cost?
• How often can the item be repaired?
• What is the lead time to get a replacement?
• What is the lead time to repair the item?
After the costs related to the stocking decision are all identified, then a logical decision can be made concerning the stocking level. For example, the difference between keeping no spares and keeping one spare could be substantial when resulting downtime is factored in. Typically critical spares have long lead times. If no spares are kept, the resulting downtime may be weeks of lost production. This can quickly add up to tens of thousands of dollars. Consider the example of a gearcase costing $24,500 that is a spare on one equipment unit.
When the gearcase is ordered, it takes 1 week to receive the spare. The holding cost is 30 percent of the price of the gearcase. While it is down, the downtime cost is $1000/hr. When it is replaced, the actual time to replace is 40 hours and the labor cost is $1000. Currently there are four spares stocked and annual usage for the past 3 years shows that one gearcase was changed each year. Based on this information, how many spares should be stocked and why?
Keeping in mind the total cost curve, you must plot the cost of investment (stocking the part) vs the cost of downtime during a failure. This can be performed in a table format or in a graph format (Fig. 7).
When the comparison is made, it is clearly seen that keeping one spare is the optimum financial decision for the company. The difference between keeping one spare (optimum) and keeping four spares (current policy) is $95,700. This is considerable savings on just one spare. Most companies have hundreds of these types of spares. Consider the savings possibilities for optimizing critical spares.
Also consider the saving potential for an organization that reduces inventory without factoring in the cost of downtime. The difference between keeping none and keeping one is $136,100. This is an even more dramatic savings. It is only when companies take this financial approach to maintenance spares stocking that inventories will truly be optimized. This is a stiff penalty to pay for short-sighted decisions.
Normal maintenance parts, supplies
The same calculations could be applied to normal stores items. If the same information is kept, statistical formulas can be used to calculate service levels, actual cost per item, number of projected turns per year, etc. These can provide a manager with the tools necessary to make accurate and cost-effective management decisions.
The drawback to all of the calculations is the need for accurate data. This is why the entire organization must have the discipline to collect accurate data before any of the techniques in this article can be utilized. Using statistical techniques with poor or inaccurate data will provide no more accurate an answer than guessing at it. A good, disciplined approach to a CMMS/EAM system will enable many of the techniques shown here to be useful to a company attempting to optimize its total asset costs.
The language of finance must always be the internal language spoken in companies today. While there may be other languages (technical, operational, etc.), ultimately the owners or shareholders want to see clear financial statements. It is only by an entire organization learning to speak the financial language that this requirement will be met. MT
Terry Wireman is senior industry analyst at GenesisSolutions, 100 Danbury Rd., Ste. 105, Ridgefield, CT 06877; (203) 431-0281
COST EFFECTIVE MAINTENANCE
Fig. 1. Decisions that involve maintenance should be based not on what is best for the operations group or
for the maintenance group, but on what is the lowest combined cost. This is the effective “bottom line” for the company.
Fig. 2. Basic Maintenance Cost Calculation
Repair cost = $1500
If the pump was serviced once every 100 hours, the cost would be $1500/100 = $15/hr
If the pump was serviced once every 500 hours, the cost would be $1500/500 = $3/hr
Service Frequency (hrs) |
Maintenance Cost ($) |
100 |
15.00 |
500 |
3.00 |
1000 |
1.50 |
1500 |
1.00 |
2000 |
0.75 |
2500 |
0.60 |
3000 |
0.50 |
3500 |
0.43 |
4000 |
0.38 |
Fig. 3. Lost Performance Calculation
If the performance loss is linear, then at 4000 hours of operation, the loss is 5 percent. If the value is $100, then the value of the loss is: 0.05 x $100/hr = $5/hr
Therefore, at 4000 hours of operation, the pump is producing a value of only $95, or it is losing $5 per hour
Time Since Last Service (hrs) |
Lost Performance Cost ($) |
100 |
0.13 |
500 |
0.63 |
1000 |
1.25 |
1500 |
1.88 |
2000 |
2.50 |
2500 |
3.12 |
3000 |
3.74 |
3500 |
4.36 |
4000 |
5.00 |
Fig. 4. True Lost Performance Calculation
The true lost performance cost is usually calculated by a calculus formula. However, a simple geometry formula can serve the purpose in this simple example.
The total loss is not the entire rectangle, but only one-half of its area. So the loss would be only one-half of the calculated amount. The revised version of the table would be: |
Time Since Last Service (hrs) |
Lost Performance Cost ($) |
|
100 |
0.065 |
||
500 |
0.31 |
||
1000 |
0.63 |
||
1500 |
0.94 |
||
2000 |
1.25 |
||
2500 |
1.56 |
||
3000 |
1.87 |
||
3500 |
2.18 |
||
4000 |
2.50 |
Fig. 5. True Total Cost Calculation
The true total cost now can be established for this task. Combining the tables in Fig. 2 and Fig. 4 results in:
Time Since Last Service (hrs) |
Maintenance Cost ($) |
Lost Performance Cost ($) |
True Total Cost ($) |
|
100 |
15.00 |
0.065 |
15.065 |
|
500 |
3.00 |
0.31 |
3.31 |
|
1000 |
1.50 |
0.63 |
2.13 |
|
1500 |
1.00 |
0.94 |
1.94 |
|
2000 |
0.75 |
1.25 |
2.00 |
|
2500 |
0.60 |
1.56 |
2.16 |
|
3000 |
0.50 |
1.87 |
2.37 |
|
3500 |
0.43 |
2.18 |
2.61 |
|
4000 |
0.38 |
2.50 |
2.8 |
Fig. 6. Factoring in Breakdown Costs
If a breakdown occurs and there is no spare, then the service also results in a cost penalty to the Operations Group:
• For planned service, 8 hr of downtime
• If a breakdown occurs, 24 hr of downtime
A breakdown will occur every 3000 hours of operation, based on repair records. The table shows
Time Since Last Service (hrs) |
Maintenance Cost ($) |
Lost Performance Cost ($) |
Downtime Cost ($) |
True Total Costs ($) |
100 |
15.00 |
0.065 |
8.00 |
23.07 |
500 |
3.00 |
0.31 |
1.60 |
4.91 |
1000 |
1.50 |
0.63 |
0.80 |
2.93 |
1500 |
1.00 |
0.94 |
0.53 |
2.47 |
2000 |
0.75 |
1.25 |
0.40 |
2.40 |
2500 |
0.60 |
1.56 |
0.32 |
2.48 |
3000 |
0.50 |
1.87 |
1.07 |
3.44 |
3500 |
0.43 |
2.18 |
0.91 |
3.52 |
4000 |
0.38 |
2.50 |
0.80 |
3.68 |
Cost of Critical Spares
Project Results |
|||
|
Total Cost |
Impact Costs |
Purchasing & Holding Costs |
0 |
$209,000.00 |
$209,000.00 |
None |
1 |
$ 72,900.00 |
$ 41,000.00 |
$ 31,900.00 |
2 |
$104,800.00 |
$ 41,000.00 |
$ 63,800.00 |
3 |
$136,700.00 |
$ 41,000.00 |
$ 95,700.00 |
4 |
$168,600.00 |
$ 41,000.00 |
$127,600.00 |
5 |
$200,500.00 |
$ 41,000.00 |
$159,500.00 |
6 |
$232,400.00 |
$ 41,000.00 |
$191,400.00 |
7 |
$264,300.00 |
$ 41,000.00 |
$223,300.00 |
8 |
$296,200.00 |
$ 41,000.00 |
$255,200.00 |
9 |
$328,100.00 |
$ 41,000.00 |
$287,100.00 |
10 |
$360,000.00 |
$ 41,000.00 |
$319,000.00 |
Fig. 7. After the costs related to a critical spares stocking situation are all identified, then a decision can be made
concerning the stocking level. Companies should plot the cost of investment vs the cost of downtim
during a failure in either a table format (above) or a graph format (below).
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